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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336

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BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336
Textbook Problem

Find the vertices, foci, and asymptotes of the hyperbola and sketch its graph.

24. 9y2 − 4x2 − 36y − 8x = 4

To determine

To Find: The vertices, foci and asymptotes of the hyperbola for the equation 9y24x236y+8x=4 .

Explanation

Given:

The hyperbola equation is as follows.

9y24x236y+8x=4 (1).

Rewrite the equation as below.

9y236y4x2+8x=49y236y+36364x2+8x+44=49y236y+364x2+8x+4=4+3649(y24y+4)4(x2+2x+1)=36

Factorize the equation as below.

9(y2)24(x+1)2=36

Divide the equation by the value 36 on both the sides.

9(y2)2364(x+1)236=36

(y2)24(x+1)29=1 (2).

Then, compare the equation (2) with the standard equation of hyperbola.

y2a2x2b2=1

Calculation:

Compute the center of the hyperbola using the equation:

(yk)2a2(xh)2b2=1(y(2))2a2(x(1))2b2=1

Therefore, the center of the hyperbola (h,k) is (1,2) .

Substitute the value of 4 for a2 and 9 for b2 in equation (2).

a2=4a=4a=2

b2=9b=9b=3

Compute the vertices.

vertices=(h,(k±a))

Substitute the value 1 for h , 2 for k and 2 for a .

vertices=(1,(2(±2)))vertices=(1,(2(2))),(1,(2(+2)))vertices=(1,0),(1,4)

Then, the vertices of the hyperbola are (1,0) and (1,4)

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